Chemistry 102
Experiment 10
_____________
Homogeneous Equilibrium: The Hydrolysis of Ethyl Acetate
In this experiment, the solutions should be made a week before they are to be used. The time for the analysis is
about four hours. In making the solutions, about half an hour is needed for measuring out the solutions, but the
heterogeneous ethyl acetate-water solution takes about an hour of intermittent shaking to prepare.
INTRODUCTION:
In this experiment, the equilibrium constant for the hydrolysis of ethyl acetate will be
determined. L. Wilhelmy was first to quantitatively measure the rate of reaction and to integrate
the rate law. This was in 1850. In 1862, Marcellin Berthelot and Péan de St. Gilles reported a
study of the hydrolysis of esters, including ethyl acetate. In their experiments, no catalyst was
used. In 1863, Guldberg and Waage interpreted this data in a theoretical way, stating their "law
of chemical equilibrium" or the law of mass action.
Keq =
[products]
[reactants]
The hydrolysis of ethyl acetate in dilute hydrochloric acid was investigated by Jones and
Lapworth. In this assignment, similar conditions will be employed. The equilibrium will be
approached from the side of ethyl acetate and water, and from the side of acetic acid and ethanol,
with 6M hydrochloric acid as a catalyst
CH3CO2--CH2CH3 + H2O
 CH3COOH + HOCH2CH3
The heat of reaction is very small; consequently, the equilibrium constant is insensitive to
temperature. For this reason, it is not necessary to control the temperature of the reaction.
PROCEDURE:
There are several parts to this experiment. The general plan is to
1. Prepare a base solution for titrating the unknown acid,
2. Standardize (determine its exact concentration) the base,
3. Titrate the equilibrium solutions with the standardized base to determine K.
Part 1: Preparation of solutions (one week prior to titration)
(1) Clean and dry five screw cap vials or test tubes with stoppers & parafilm. Number
them 1, 2, 3, 4, and 5.
(2) Table 1 summarizes the instructions below. Clean a 5-mL volumetric pipette by
rinsing it a few times with distilled water from a squeeze bottle. Using this pipette,
add a 5-mL aliquot of distilled water to test tubes labeled 1 & 2. DO NOT USE YOUR
MOUTH in this experiment to fill the pipette; use the pipette bulb.
(3) Rinse the pipette with 6M HCI.
(4) Using densities of the liquids and solutions given in handbooks, and the nominal
volumes, the weights of each reactant can be obtained
density x volume = weight.
(5) After rinsing the pipette with 6M hydrochloric acid, add 5-mL aliquots of 6M
hydrochloric acid to Tubes 1-5. Rinse the pipette with distilled water then acetone
from a squeeze bottle and allow to air dry.
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Chemistry 102
(6) Add a 5-mL aliquot of ethyl acetate to Tube 3. Use care when handling ethyl acetate
since it is volatile.
(7) Using a clean and dry 5-mL graduated Mohr pipette, add a 3.0-mL aliquot of pure
acetic acid to Tube 4 and a 2.0-mL aliquot to Tube 5.
(8) With a clean Mohr pipette, add a 2.0-mL aliquot of ethyl alcohol to Tube 4 and a 3.0ml aliquot to Tube 5.
(9) Close all the tubes tightly immediately after adding the samples. Shake the tubes
well. Shake the heterogeneous mixtures for a few minutes. Let the tubes stand for a
week, if possible, shaking them now and then.
Table 1 summarizes the above instructions
Tube or
Water
6M HCI in
vial #
H20
Ethyl Acetate
(EtOAc)
Acetic acid
(HOAc)
Ethanol
(EtOH)
1
5.0 mL
5.0 mL
0
0
0
2
5.0 mL
5.0 mL
0
0
0
3
0
5.0 mL
5.0 mL
0
0
4
0
5.0 mL
0
3.0 mL
2.0 mL
5
0
5.0 mL
0
2.0 mL
3.0 mL
Part 2: Standardization of a 2M solution of Sodium Hydroxide:
(1). Preparation of a Base Solution: It will be necessary to prepare 500 mL of an
approximately 2.0 M NaOH. This should be done by diluting a 6.0 M shelf reagent. Make the
necessary calculations using VcMc = VdMd. Remember to keep this solution for part 3.
(2). Standardization of the Base: Shelf reagents are not prepared to any high degree of
accuracy, and even if they were, could not be kept long at a particular concentration due to
gradual changes from exposure. If a solution is prepared from a shelf reagent by dilution, no
matter how carefully measurements are made, the result will be a solution of undetermined
concentration. It must be standardized to know its concentration beyond two significant figures.
Use oxalic acid, H2C2O4∙2H2O, as the acid standard. Weigh out exactly about 2 to 3 g grams of
oxalic acid on an analytical balance and record the weight to five significant figures. Use
weighing by difference technique, that is, weight of flask plus acid minus the weight of flask, to
obtain the weight of acid. (Remember, at this accuracy level, fingerprints cause errors.) Place the
oxalic acid in a clean 125 or 250 mL Erlenmeyer flask, and dissolve it in 25 to 30 mL of distilled
water. Titrate the acid solution with the NaOH solution from a clean 50 mL burette. Using
phenolphthalein as an indicator, titrate to a light pink color. Determine very precisely the volume
of base required to reach the endpoint by approximating between the marks on the burette for a
fourth significant figure. A proper standardization requires as a minimum three trials that agree
within plus or minus 0.5% of each other. Repeat the standardization process until three trials do
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Chemistry 102
agree within these limits. If there is difficulty staying within the limits, the most likely reasons
are:
1.
2.
3.
4.
5.
failure to read the burette accurately,
unclean equipment,
not titrating to a consistent indicator color,
splashing of reagents, and/or
errors in weighing.
In your lab notebook, record the following table and show all calculations.
Trial
1
2
3
Mass of flask
Mass of flask & oxalic
acid
Mass of oxalic acid
Moles of oxalic acid
Moles of NaOH
Volume of NaOH titrated
Molarity of NaOH
Part 3: Analysis of equilibrium solutions- one week later:
(1). Open a tube (or vial) and pour its contents into a 125-mL Erlenmeyer flask. Rinse the
tube with distilled water from a squeeze bottle, no more than 20 mL. Add a few drops of
bromothymol blue (or phenolphthalein) indicator solution.
(2). Titrate the acid with the standardized sodium hydroxide from part 2. Swirl
continuously as you titrate to the first sign of color change. The end point will fade in a few
seconds, however, because of the reaction of ethyl acetate with sodium hydroxide, which
uses up hydroxide ion
EtOAc + OH-  -OAc + EtOH
Consequently, titrate quickly, not dropwise. Do not try to get a permanent end point. Treat all of
the tubes in this way.
Equipment and supplies
- One 5-ml volumetric pipette
- One 5-ml Mohr graduated pipette
- Five 150-mm test tubes and corks
- Two 125-ml Erlenmeyer flasks
- 10 -ml Glacial acetic acid
- 10 ml Ethyl acetate
- 150 ml 6M Sodium hydroxide
- 1 mL Phenolphthalein indicator
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- One pipette bulb
- One 50-ml burette
- Plastic squeeze bottle
- Analytical balance
- 10 ml Ethyl alcohol, absolute
- 30 ml 6M Hydrochloric acid
- 2 to 3 g Oxalic acid
- Acetone in squeeze bottle
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Chemistry 102
CALCULATIONS:
(1) Calculate the initial number of moles, written as no, of the reactants ethyl acetate
(EtOAc), ethyl alcohol (EtOH), and acetic acid (HOAc) in Tubes 3, 4, and 5 at the beginning
(time equals zero or t=0), from the weights of the aliquot and the molecular weights (d = m/V &
MW = m/n). The weight of an aliquot is obtained by multiplying known densities by the
nominal volume of the aliquot. The density at 20oC is 1.098 g/mL for 6M hydrochloric acid;
0.901 g/mL' for ethyl acetate; 1.049 g/mL for acetic acid; and 0.7893 g/mL for ethyl alcohol. The
chemical formulas are ethyl acetate=CH3COOCH2CH3; acetic acid=CH3COOH; and ethyl
alcohol=CH3CH2OH.
For example: (ethyl acetate, vial 3)
5.0 mL (0.901 g/mL) = 4.505 g ethyl acetate
4.505 g / 88g/mol = 0.05119 mol ethyl acetate
(2) The number of moles of water present initially, no (H2O) in Tubes 3, 4, and 5 can be
calculated from the weight of water present in the 5.0-mL aliquot of 6M HCI. Theoretically, the
weight of water present in the 5-mL aliquot of 6M HCI is equal to the weight of the 6M solution
less the weight of hydrogen chloride in it
(c) no (H2O) = msoln - mHCl
18 g/mol
The msoln can easily be determined from the given density and measured initial volume (Vd=m)
but how will the mass of HCl present be determined? This is the reason for doing a titration.
The weight of hydrogen chloride can be calculated from the number of moles of HCI, which can
be determined from the titration of test tubes 1 & 2 with sodium hydroxide.
(a) nHCl = VNaOH MNaOH
then (b) mHCl = (nHCl ) (36.53 g/mol)
(3) The number of moles of acetic acid present at equilibrium, nHOAc, in Tubes 3, 4, and 5
is equal to the total number of moles of acid in the tube minus the number of moles of HCI in the
tube. The total acid is determined by titrating the equilibrium solution in the tube
(a) total moles of acid = VNaOH MNaOH
(b) nHOAc = (total moles of acid) - nHCl (from step 2a)
(4a) In Tube 3, only ethyl acetate and water are present initially with the HCI catalyst. The
reaction for attaining equilibrium can be written:
CH3CO2CH2CH3 + H2O  CH3COOH + HOCH2CH3
The number of moles of ethyl acetate left at equilibrium (nEtOA) will be equal to the number at the
beginning (no EtOAc) minus the number of moles of acetic acid found at equilibrium (nHOAc from
step 3) in Tube 3:
nEtOA = (noEtOAc) - (nHOAc)
This is because a mole of acetic acid is formed by the reaction of one mole of ethyl acetate with
one mole of water.
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Chemistry 102
(4b) The number of moles of ethanol must be the same as that of acetic acid because,
according to the equation, each mole of ethyl acetate that reacts with one mole of water makes
one mole of acetic acid and one mole of ethanol.
nEtOH = nHOAc
(4c) The number of moles of water is: nH2O = noH2O - nHOAc
(step 1) (step 3)
(5) In Tubes 4 and 5, the reverse reaction takes place
CH3COOH + HOCH2CH3  CH3COOCH2CH3 + H2O
Consequently, the number of moles of acetic acid originally present minus the number found at
equilibrium, is the number of moles of ethyl acetate present at equilibrium. This number is also
the number of moles of water formed by the reaction. The number of moles of ethanol is the
number of moles of ethanol initially present, less the number of moles of acetic acid that reacted
with the ethanol making ethyl acetate.
nEtOAc
=
noHOAc
-
nHOAc
nH2O
=
noH2O
+
nEtOAc
nEtOH
=
noEtOH
-
nEtOAc
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Chemistry 102
NAME_____________________________________
SECTION _______
REPORT SHEET
EXPERIMENT 10
Average Molarity of NaOH:_____________________
Homogeneous Equilibrium: The Hydrolysis of Ethyl Acetate
Table 2: Summary of data calculated in steps 1-5
Test tube 3
Test tube 4
Test tube 5
no (EtOAc)
no (EtOH)
no (HOAc)
no (H2O)
n (HCl)
n (HOAc)
n (EtOH)
n (EtOAc)
n (H2O)
(6) The equilibrium expression for the equilibrium
CH3CO2CH2CH3 + H2O  CH3COOH + HOCH2CH3
is written as
Keq =
[CH3CH2OH]
[CH3COOH] = [EtOH] [HOAc]
[CH3COOCH2CH3] [H2O]
[EtOAc] [H2O]
Since, in a particular tube, the volume of the solution is the same for all quantities, Keq can
be written
Keq =
nEtOH nHOAc
nEtOAc nH2O
where n is the number of moles of the components at equilibrium. Calculate Keq for each
equilibration that you performed. Average the values from each equilibration, where the
catalyst concentration was 6M.
Keq = _____________ , _______________, _______________
Average Keq = __________________
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